Answer to Question #179028 in C++ for simisola oluwaseyitan alabi

Question

Assume a disk with the following characteristics: Number of sectors per track = 20 Number of tracks per surface = 50 Number of surfaces = 2 (called a double-sided disk) Number of characters per sector = 1,024 Arm movement time = 0.4 msec to move 1 track in any direction Rotation speed = 2,400 rev/min

1. How many characters can be stored on this disk?

2. What are the best-case, worst-case, and average-case access times for this disk? (Assume that the average seek operation must move 20 tracks.)

Answer 1

1. The total number of characters (ch) is 2 surfaces/disk 3 50 tracks/surface 3 20 sectors/track 3 1024 ch/sector which is 2,048,000 characters on a single disk.

2. The seek time depends on the number of tracks over which the read head must move. This could range from 0, if the arm does not need to move, to a worst case of the arm having to move from the far inside track to the far outside track, a total of 49 tracks. The average, as stated in the problem, is a move across 20 tracks. The best-case rotational delay is 0, whereas the worst case is one complete revolution. The rotational speed is 2400 rev/min = 40 rev/sec = 25 msec/rev. On the average we will wait about 1/2 a revolution. Finally, the transfer time is the same in all cases, the time it takes for one sector (1/20 of a track) to rotate under the read/write head, which is 1/20 rev * 25 msec/rev = 1.25 msec. Putting all this together in a table produces the following values for the time (in msec) required for each task: Best Case Average Case Worst Case Seek time 0.0 20 * 0.4 = 8.0 49 * 0.4 = 19.6 Latency 0.0 0.5 * 25 = 12.5 1 * 25 = 25.0 Transfer 1.25 1.25 1.25 Total 1.25 21.75 45.85